KCET · Physics · Alternating Current
A person wants a real image of his own, \( 3 \) times enlarged. Where should he stand infront of a concave mirror of radius of curvature \( 30 \mathrm{~cm} \) ?
- A \( 10 \mathrm{~cm} \)
- B \( 30 \mathrm{~cm} \)
- C \( 90 \mathrm{~cm} \)
- D \( 20 \mathrm{~cm} \)
Answer & Solution
Correct Answer
(D) \( 20 \mathrm{~cm} \)
Step-by-step Solution
Detailed explanation
Given, magnification \( =3 \); radius of curvature \( =30 \mathrm{~cm} \)
Using mirror formula, \( \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \)
Now \( m=\frac{-v}{u} \Rightarrow-3=\frac{-v}{u} \Rightarrow v=3 u \)
Also \( f=\frac{R}{2}=\frac{30}{2}=15 \mathrm{~cm} \)
Therefore,
\( \frac{1}{-15}+\frac{1}{3 v}+\frac{1}{u} \Rightarrow \frac{-1}{15}=\frac{4}{3 u} \Rightarrow-3 u=60 \) \( \Rightarrow u=-20 \mathrm{~cm} \)
Thus, the e-ser
Using mirror formula, \( \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \)
Now \( m=\frac{-v}{u} \Rightarrow-3=\frac{-v}{u} \Rightarrow v=3 u \)
Also \( f=\frac{R}{2}=\frac{30}{2}=15 \mathrm{~cm} \)
Therefore,
\( \frac{1}{-15}+\frac{1}{3 v}+\frac{1}{u} \Rightarrow \frac{-1}{15}=\frac{4}{3 u} \Rightarrow-3 u=60 \) \( \Rightarrow u=-20 \mathrm{~cm} \)
Thus, the e-ser
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